if h0=0, then
c=−g2x2+v0x = -x(g2x-v0)
clearly there are two solutions:
x=0
x = v0/g2
The equation c=−g2x2+v0x+h0 gives the instantaneous height, c, of a projectile based on a given real-world scenario. If h0=0, for what values of c is there exactly one solution for x? Explain.(1 point)
a ) There is exactly one solution for x only when c is zero. Since the equation is quadratic, there will always be two solutions except for the point where the projectile strikes the ground.
b ) There are no values of c where there is exactly one solution for x. Since the equation is quadratic, there will always be either two solutions or no solutions.
c )There is exactly one solution for x only when c is equal to the maximum height. Since the initial height is zero, the projectile reaches all heights in its trajectory (except the peak) twice—once on the way up and once on the way down.
d) There is exactly one solution for x only when c is zero or when c is equal to the maximum height. Since the equation is quadratic, there will always be two solutions except for the points where the projectile reaches its peak and where it strikes the ground.
2 answers
so what im getting is... There are no values of c where there is exactly one solution for x. Since the equation is quadratic, there will always be either two solutions or no solutions.
2 answers